series formula

StatusX

I'm trying to get an expression for the nth term in this series:

[tex]c_1 = 1[/tex]

[tex]c_n = \sum_{k=1}^{n-1} \frac{c_n}{(n-k)!}[/tex]

For example:

[tex]c_2 = 1/1! = 1[/tex]

[tex]c_3 = 1/2! +1/1! = 3/2[/tex]

[tex]c_4 = 1/3! + 1/2! + 3/(2\cdot 1!) = 13/6[/tex]

etc.

Since the factor in front of each term is different in each series, you can't express [itex]c_n[/itex] just in terms of [itex]c_{n-1}[/itex]. I have no idea how to start.

StatusX

First, that is supposed to be c_k inside the sum. It's too late to edit it.

Is no one replying because this is too hard or too easy? If it helps, this was found when I was trying to derive a formula for

[tex]\sum_{n=1}^{\infty} \frac{1}{n^{2k}} [/tex]

I know this involves the Bernoulli numbers, so maybe they are involved in the series from my first post.

Data

[tex] \sum_{n=1}^\infty \frac{1}{n^{2k}} = \zeta(2k), \; \forall k > \frac{1}{2}[/tex]

http://mathworld.wolfram.com/RiemannZetaFunction.html

Is there a particular [itex]k[/itex] that you want to evaluate it for?

StatusX

I was using fourier series to find an explicit formula. Namely, I started with a sawtooth wave with fourier coefficients proportional to 1/n. Then I normalized it so the integral over one period was one, which meant that:

[tex]\sum_{n=1}^{\infty} |c_n|^2 = 1[/tex]

From which I got the first sum (k=1). Then I integrated to get a series with 1/n²,l fixed the constant so there were only sinusoidal terms in the fourier series, and renormalized. I correctly got the formula for k = 1, 2, and 3. Then trying to find the general form, which is made complicated by the integrating constants that depend on the constants from previous iterations, I ran into a series very similar to the one from the first post. I was just curious if there is a closed form for that expression, or if you have to resort to more complicated methods.

damoclark

Maybe this will help, although I don't surely know cause I haven't tried it yet, but I recall a method I used to solve a similar problem once.

Set:

[tex] f(x) := c_1 + c_2x + c_3x^2 + c_4x^3 + c_5x^4 + .....[/tex]

where [tex]c_k[/tex] are the co-efficients you originally defined.


Then expand the inverse of [tex]f(x)[/tex] as a power series, call this new power series [tex]g(x)[/tex], say. I'm assuming the co-efficients of [tex]g(x)[/tex] might be easier to recognise. If you have maple or mathematica this can be easily done.

Then the [tex] c_n [/tex] co-efficients of [tex]f(x)[/tex] can be calculated by knowing the n'th derivitive of [tex]1/g(x)[/tex] , evaluated at the point zero.

If you can calculate it, I'd be interested to know what the power series of [tex]g(x)[/tex] is.